sin2t+sin2(pi)t的平方积分

Question

sin2t+sin2(pi)t的平方积分

Answer 1

∫(sin2t+sin2πt)^2dt
= ∫[(sin2t)^2+2sin2tsin2πt+(sin2πt)^2]dt
= (1/4)∫{(1-cos4t)+4[cos2(π-1)t-cos2(π+1)t]+(1-cos4πt)}dt
= (1/4)∫[2-cos4t-cos4πt+4cos2(π-1)t-4cos2(π+1)t]dt
= t/2-(1/16)sin4t-[1/(16π)]sin4πt
+[1/(2π-2)]sin2(π-1)t-[1/(2π+2)]sin2(π+1)t+C.